consequently : 
cos* 20 — 2(1,02875) cos 20 + 0,58074 — 0 
cos 20 = 1,02875 + V(1,02875)? — 0,58074 = 
= 1,02875 + 0,68948 — 1,71823 or 0.33927 
Only the second value corresponds, so that: 
20 = 70°10" or — 70°10! 
@ == 5075) OE — in kag aa 
a 0,43625 0.3609€ 
sin 6 = — — cot vo = — —— cot o = 0.36096 cot o 
ri i —1,20858 \ ; js 
from which 
I= De = ae ae. 
0, = — 35°5' , o, = — 30°53’. 
As will appear afterwards the optic extinction offers an expedient 
to decide in a given case whether one has to do with the secant 
plane (9,6,) or with (@,6,). 
In a graphical way the problem of the orientation of crystal- 
sections can be solved in a considerably simpler way. To do so one 
can make use of diagrams, that give for any discretional angle 
between two planes, the apparent angle / as « funetion of @ and 6. 
Fig. 3. 
In fig. 3 be £ again the projection plane, V the erystalplane, 
